Match List I with List II : List I List II A Isothermal Process I Work done by the gas decreases internal energy B Adiabatic Process II No change in internal energy C Isochoric Process III The heat ab

Question

Match List I with List II : List I List II A Isothermal Process I Work done by the gas decreases internal energy B Adiabatic Process II No change in internal energy C Isochoric Process III The heat absorbed goes partly to increase internal energy and partly to do work D Isobaric Process IV No work is done on or by the gas Choose the correct answer from the options given below :

TestHub · Accepted Answer

(A) Change in internal energy is expressed as $∆ U = n C_{v} ∆ T$ , here, $∆ T$ is change in temperature.

Since, in an isothermal process temperature remains constant, thus, $∆ U = 0$

$A \to I I$

(B) In an adiabatic process, heat transfer, $Q = 0$ .

So, from first law of thermodynamics,

$Q = ∆ U + W$

$∆ U = - W$

Since, work done by gas is positive, thus, $∆ U$ is negative

$B \to I$

(C) In an isochoric process, volume remains constant, so work done by or on the gas $W = P ∆ V = 0$

$C \to I V$

(D) In an isobaric process, pressure remains constant, so work done $W = P ∆ V \neq 0$ and change in internal energy $∆ U = n C_{v} ∆ T \neq 0$

Thus, heat absorbed goes partly to increase internal energy and partly do work.

$D \to I I I$

Physics - Heat & Thermo Question with Solution | TestHub

Options:

Doubts & Discussion

	List I		List II
A	Isothermal Process	I	Work done by the gas decreases internal energy
B	Adiabatic Process	II	No change in internal energy
C	Isochoric Process	III	The heat absorbed goes partly to increase internal energy and partly to do work
D	Isobaric Process	IV	No work is done on or by the gas